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Equivalent capacitance; complex circuit |
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| Nov5-08, 03:03 PM | #1 |
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Equivalent capacitance; complex circuit
1. The problem statement, all variables and given/known data
Hello, I was given this problem as a homework assigment. Each capacitor in the figure has capacitance C. What is the equivalent capacitance between points a and b? The answer is given in "C_eq/C" ![]() 2. Relevant equations I used the "C_eq={(1/C_1)+(1/C_2)+.....(1/C_n)}^-1" for the capacitors in series. I used the "C_eq=(C_1)+(C_2)+.....(C_n) for the capacitors in parallel. 3. The attempt at a solution Combining everything that was in parallel and everything that was in series until it come down to one capacitor, the answer that i get is C or 1. Any ideas on what I am doing wrong or if im inputing the answer in the wrong way? |
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| Nov5-08, 06:15 PM | #2 |
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Welcome to PF.
It may be just me, but I don't see any series capacitors in this circuit. This one may require Kirchoff's loop and node laws (also known as KVL and KCL). |
| Nov5-08, 08:50 PM | #3 |
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I think it is 2C.
If you consider capacitors like resistances, you will see bridge and one of the capacitor behaves as open circuited. |
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| complex circuit, equivalent capacitor, parallel, series |
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